Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-2x-3y &= -8 \\ -8x+9y &= 4\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $-8x = -9y+4$ Divide both sides by $-8$ to isolate $x$ $x = {\dfrac{9}{8}y - \dfrac{1}{2}}$ Substitute this expression for $x$ in the first equation. $-2({\dfrac{9}{8}y - \dfrac{1}{2}}) - 3y = -8$ $-\dfrac{9}{4}y + 1 - 3y = -8$ Simplify by combining terms, then solve for $y$ $-\dfrac{21}{4}y + 1 = -8$ $-\dfrac{21}{4}y = -9$ $y = \dfrac{12}{7}$ Substitute $\dfrac{12}{7}$ for $y$ in the top equation. $-2x-3( \dfrac{12}{7}) = -8$ $-2x-\dfrac{36}{7} = -8$ $-2x = -\dfrac{20}{7}$ $x = \dfrac{10}{7}$ The solution is $\enspace x = \dfrac{10}{7}, \enspace y = \dfrac{12}{7}$.